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Thirty two persons $X_{1}, X_{2}, \ldots, X_{32}$ are randomly seated around a circular table at equal intervals. Two persons $X_{i}$ and $X_{j}$ are said to be within earshot of each other if there are at most three persons between them on the minor arc joining $X_{i}$ and $X_{j}$. The probability that $X_{1}$ and $\mathrm{X}_{2}$ are within earshot of each other is, $\left(\right.$ Here $\left.\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) ! \mathrm{r} !}\right)$
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Verified Answer
The correct answer is:
$\frac{8}{31}$
Case - 1
No person between $\mathrm{X}_{1} \& \mathrm{X}_{2}$
$\frac{30 ! \times 2 !}{31 !}=\frac{2}{31}$
Case-2
One person between $X_{1} \& X_{2}$
$\frac{{ }^{30} \mathrm{C}_{1}(29) ! \times 2 !}{31 !}=\frac{2}{31}$
Case-3
When 2 person between $\mathrm{X}_{1} \& \mathrm{X}_{2}$
$\frac{{ }^{30} \mathrm{C}_{2} \times 28 ! \times 2 ! \times 2 !}{31 !}=\frac{2}{31}$
Case-4
When 3 person between $X_{1} \& X_{2}$ $\frac{{ }^{30} \mathrm{C}_{3} \times 27 ! \times 2 ! \times 3 !}{31 !}=\frac{2}{31}$
Total $=\frac{8}{31}$
No person between $\mathrm{X}_{1} \& \mathrm{X}_{2}$
$\frac{30 ! \times 2 !}{31 !}=\frac{2}{31}$
Case-2
One person between $X_{1} \& X_{2}$
$\frac{{ }^{30} \mathrm{C}_{1}(29) ! \times 2 !}{31 !}=\frac{2}{31}$
Case-3
When 2 person between $\mathrm{X}_{1} \& \mathrm{X}_{2}$
$\frac{{ }^{30} \mathrm{C}_{2} \times 28 ! \times 2 ! \times 2 !}{31 !}=\frac{2}{31}$
Case-4
When 3 person between $X_{1} \& X_{2}$ $\frac{{ }^{30} \mathrm{C}_{3} \times 27 ! \times 2 ! \times 3 !}{31 !}=\frac{2}{31}$
Total $=\frac{8}{31}$
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